Wavelet admissibility This question is about Daubechies' book "Ten Lectures on Wavelets", section 2.4.
This author says, 
if $\psi \in L^2(\mathbf{R})$, $\int_{\mathbf{R}}dx\ \psi(x) = 0$, and
\begin{align*}
\exists \alpha>0 \quad \int_{\mathrm{R}}dx (1+|x|)^\alpha\ |\psi(x)| < \infty
\end{align*}
holds, then $|\hat{\psi}(\xi)| \leq C|\xi|^\beta$ follows, where 
$\beta = \min(\alpha, 1)$, 
$C$ is some constant, and $\hat{\psi}$ is the Fourier transform of $\psi$.
And he says that from this results, 
\begin{align*}
2\pi \int d\xi \ |\xi|^{-1} |\hat{\psi}(\xi)|^2 < \infty
\end{align*}
follows.
I cant't think of how to derive the former result. 
And, I think the latter result doesn't follow from the former, because
\begin{align*}
2\pi \int d\xi \ |\xi|^{-1} |\hat{\psi}(\xi)|^2 \leq 
2\pi C^2 \int d\xi \ |\xi|^{2\beta-1} 
\end{align*}
and the right hand side does't have any upper bound if, say, $\alpha = 2$.
 A: From the conditions on $\psi$ we have $$
|\psi(\xi)|=\Bigl|\int_\mathbb{R}\psi(x)(e^{-ix\xi}-1)\,dx\Bigr|\le\Bigl(\int_\mathbb{R}|\psi(x)|(1+|x|)^\alpha\,dx\Bigr)\sup_{x}\frac{|e^{-ix\xi}-1|}{(1+|x|)^\alpha}.
$$
If $\alpha\ge1$, it is clear that
$$\frac{|e^{-ix\xi}-1|}{(1+|x|)^\alpha}\le|\xi|.$$
For the case $0<\alpha<1$, observe that $|e^{-ix\xi}-1|$ is periodic of period $\pi/|\xi|$ as a function o f $x$. Then, since $1/(1+x)^\alpha$ is decreasing, $$\sup_{x}\frac{|e^{-ix\xi}-1|}{(1+|x|)^\alpha}\le\sup_{0\le x\le\pi/|\xi|}\frac{|e^{-ix\xi}-1|}{(1+|x|)^\alpha}\le\sup_{0\le x\le\pi/|\xi|}\frac{x\,|\xi|}{(1+x)^\alpha}.$$
Since $x/(1+x)^\alpha$ is decreasing (because $\alpha<1$),
$$\sup_{0\le x\le\pi/|\xi|}\frac{x\,|\xi|}{(1+x)^\alpha}=\frac{\pi\,|\xi|^\alpha}{(|\xi|+\pi)^\alpha}\le\pi^{1-\alpha}\,|\xi|^\alpha.$$
As for the second estimate, are you sure that there is no further condition on $\psi$? If for instance $\psi\in L^2$, then
$$
\int_\mathbb{R}\frac{|\hat\psi(|\xi)|^2}{|\xi|}\,d\xi\le C\int_{|\xi|\le1}|\xi|^{-1+2\beta}\,d\xi+\int_{|\xi|>1}|\hat\psi(\xi)|^2\,d\xi<\infty.
$$
